A gray–iron casting product is been molded with a green sand stacking molding process; which is known to produce pin–hole defects. The current yield is 76%. The plant manager wants to increase the plant’s yield.
Pouring molten metal into the ladle
The pin–holes are associated with the molten metal temperature at the furnace, at the time of the pouring process. The acceptable minimum is 1127 C. However, the gray iron alloy is known to be stable at 1204 C.
Pouring molten metal into the stack of molds
An experiment was conducted to verify
\[yield=f(temp)+\epsilon\] at three temperatures. The runs’ results are shown in table 1.
| 1127 | 1178 | 1204 |
|---|---|---|
| 73.4 | 74.4 | 78.9 |
| 76.3 | 74.2 | 79.3 |
| 76.9 | 72.7 | 77.9 |
| 77.9 | 73.5 | 78.9 |
| 77.7 | 71.4 | 79.5 |
data.set <- data.frame(yield=
c(73.4,76.3,76.9,77.9,77.7,
74.4,74.2,72.7,73.5,71.4,
78.9,79.3,77.9,78.9,79.5),
temp=as.factor(
c(rep('1127',5),
rep('1178',5),
rep('1204',5)
)))
data.set
## yield temp
## 1 73.4 1127
## 2 76.3 1127
## 3 76.9 1127
## 4 77.9 1127
## 5 77.7 1127
## 6 74.4 1178
## 7 74.2 1178
## 8 72.7 1178
## 9 73.5 1178
## 10 71.4 1178
## 11 78.9 1204
## 12 79.3 1204
## 13 77.9 1204
## 14 78.9 1204
## 15 79.5 1204
#Boxplot
fn <- yield~temp
boxplot (fn,data.set)
library(gplots)
plotmeans(fn,data.set,ci.label=T,n.label=F)
bartlett.test(fn,data.set)
##
## Bartlett test of homogeneity of variances
##
## data: yield by temp
## Bartlett's K-squared = 3.6215, df = 2, p-value = 0.1635
fligner.test(fn,data.set)
##
## Fligner-Killeen test of homogeneity of variances
##
## data: yield by temp
## Fligner-Killeen:med chi-squared = 1.9389, df = 2, p-value = 0.3793
The potential relation of the yield and the temperature are explored as a fixed effect model:
\[Y_{ij}=\mu+\tau_i+\epsilon_{ij}\]
Where:
fit_1 <- aov(fn,data.set)
summary(fit_1)
## Df Sum Sq Mean Sq F value Pr(>F)
## temp 2 80.55 40.27 23.32 7.34e-05 ***
## Residuals 12 20.72 1.73
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
TukeyHSD(fit_1)
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = fn, data = data.set)
##
## $temp
## diff lwr upr p adj
## 1178-1127 -3.20 -5.4173783 -0.9826217 0.0060433
## 1204-1127 2.46 0.2426217 4.6773783 0.0297891
## 1204-1178 5.66 3.4426217 7.8773783 0.0000517
power.anova.test(groups = 3, n = 5, between.var = 40.27, within.var = 1.73, power = NULL)
##
## Balanced one-way analysis of variance power calculation
##
## groups = 3
## n = 5
## between.var = 40.27
## within.var = 1.73
## sig.level = 0.05
## power = 1
##
## NOTE: n is number in each group
require(margins)
cplot(fit_1)
## xvals yvals upper lower
## 1 1127 76.44 77.59189 75.28811
## 2 1178 73.24 74.39189 72.08811
## 3 1204 78.90 80.05189 77.74811
## $grandsum
## Grandmean df.bet df.with MS.bet MS.with
## 76.19 2.00 12.00 40.27 1.73
## F.stat F.prob SS.bet/SS.tot
## 23.32 0.00 0.80
##
## $stats
## Size Contrast Coef Wt'd Mean Mean Trim'd Mean Var. St. Dev.
## X1178 5 -2.95 73.24 73.24 73.47 1.50 1.23
## X1127 5 0.25 76.44 76.44 76.97 3.30 1.82
## X1204 5 2.71 78.90 78.90 79.03 0.38 0.62